How is this multiaxial relationship derived in this paper?

How did the authors arrive at $$\frac{3}{2}\left(\frac{S_{i j}}{\sigma_{\mathrm{e}}}\right)$$ in the second equation below? Chain rule is obvious but, I can't get the first term.

With the consideration of initial yield stress, $$k$$, and ignoring the work hardening and other state variables, energy dissipation potential can be in the form of $$\psi=\frac{K}{n+1}\left(\frac{\sigma_{\mathrm{e}}-k}{K}\right)^{n+1}\tag{1}$$ Where $$\sigma_{\mathrm{e}}=\left(3 S_{i j} \cdot S_{i j} / 2\right)^{1 / 2}$$ is the effective stress, $$S_{i j}=$$ $$\sigma_{i j}-\delta_{i j} \sigma_{k k} / 3$$ is the component of stress deviator tensor (the Einstein summation convention of summing on repeated indices is used in this paper), $$\sigma_{i j}$$ is the component of stress tensor and $$\delta_{i j}$$ is the kronecker delta. $$K$$ and $$n$$ are material constants. Assuming normality and the associated flow rule, the multiaxial relationship is given by $$\frac{\mathrm{d} \varepsilon_{i j}^{\mathrm{p}}}{\mathrm{d} t}=\frac{\partial \psi}{\partial S_{i j}}=\frac{3}{2}\left(\frac{S_{i j}}{\sigma_{\mathrm{e}}}\right)\left(\frac{\sigma_{\mathrm{e}}-k}{K}\right)^{n}\tag{2}$$ Where $$\varepsilon_{i j}^{\mathrm{p}}$$ is the component of plastic strain tensor.

The paper referred here:

Chen, Y., Zhuang, W., Wang, S., Lin, J., Balint, D., & Shan, D. (2012). Investigation of FE model size definition for surface coating application. Chinese Journal of Mechanical Engineering, 25(5), 860-867.


Taking into account that $$\sigma_e=\left(\frac{3}{2}S_{ij}S_{ij}\right)^{1/2}$$ then $$\frac{\partial\sigma_e}{\partial S_{hk}}= \frac{1}{2}\left(\frac{3}{2}S_{ij}S_{ij}\right)^{-1/2}\frac{\partial}{\partial S_{hk}}\frac{3}{2}S_{ij}S_{ij}= \frac{1}{2\sigma_e}\cdot\frac{3}{2}\frac{\partial}{\partial S_{hk}}S_{ij}S_{ij}= \frac{1}{2\sigma_e}\cdot\frac{3}{2}2S_{hk}= \frac{3}{2}\left(\frac{S_{hk}}{\sigma_e}\right)$$